Combinations

We know that:

ⁿC_r

⇒²⁰C₄

⇒²⁰C₄

⇒²⁰C₄

⇒²⁰C₄

⇒²⁰C₄

⇒²⁰C₄

⇒²⁰C₄ =4845

Ans: ²⁰C₄ =4845

We know that:

ⁿC_r =

⇒¹⁶C₁₃=

⇒¹⁶C₁₃=

⇒¹⁶C₁₃ =

⇒¹⁶C₁₃ =

⇒¹⁶C₁₃ =

⇒¹⁶C₁₃ =

⇒¹⁶C₁₃ =560

Ans: ¹⁶C₁₃=560

We know that:

ⁿC_r =

⇒⁹⁰C₈₈=

⇒⁹⁰C₈₈=

⇒⁹⁰C₈₈=

⇒⁹⁰C₈₈=

⇒⁹⁰C₈₈=

⇒⁹⁰C₈₈=

⇒⁹⁰C₈₈=4005

Ans: ⇒⁹⁰C₈₈=4005

We know that:

ⁿC_r =

⇒⁷¹C₇₁=

⇒⁷¹C₇₁=

⇒⁷¹C₇₁= …

⇒⁷¹C₇₁=1

Ans: ⁷¹C₇₁=1

We know that:

ⁿC_r =

⇒ⁿ⁺¹C_n=

⇒ⁿ⁺¹C_n=

⇒ⁿ⁺¹C_n= …

⇒ⁿ⁺¹C_n=

⇒ⁿ⁺¹C_n=

⇒ⁿ⁺¹C_n=n+1

Ans: ⁿ⁺¹C_n=n+1

We know that:

Ans:

(i) Given: ¹⁵C₈ + ¹⁵C₉ – ¹⁵C₆ – ¹⁵C₇

To prove: ¹⁵C₈ + ¹⁵C₉ – ¹⁵C₆ – ¹⁵C₇=0

We know that:

ⁿC_r = ⁿC_n-r

⇒

⇒¹⁵C₈ + ¹⁵C₉ – ¹⁵C₆ – ¹⁵C₇=¹⁵C₈ + ¹⁵C₉ – ¹⁵C₉ – ¹⁵C₈=0

Hence,proved that ¹⁵C₈ + ¹⁵C₉ – ¹⁵C₆ – ¹⁵C₇=0

(ii) We know that: ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r

Therefore, n = 10 and r = 4

L.H.S = ¹⁰C₄ + ¹⁰C₃ = ¹¹C₄

Hence, proved.

(i) Given: ⁿC₇ = ⁿC₅

To find:n=?

We know that:

ⁿC_r = ⁿC_n-r

⇒ ⁿC₇= ⁿC_n-7

⇒ ⁿC_n-7= ⁿC₅

⇒n-7=5

⇒n=7+5=12

Ans:n=12

(ii) Given: ⁿC₁₄ = ⁿC₁₆

To find: ⁿC₂₈=?

We know that:

ⁿC_r = ⁿC_n-r

⇒ ⁿC₁₄= ⁿC_n-14

⇒ ⁿC_n-14= ⁿC₁₆

⇒n-14=16

⇒n=16+14=30

⇒n=30

So,

ⁿC₂₈= ³⁰C₂₈

⇒³⁰C₂₈=

⇒³⁰C₂₈=

⇒³⁰C₂₈=

⇒³⁰C₂₈=

⇒³⁰C₂₈=

⇒³⁰C₂₈=435

Ans: ³⁰C₂₈=435

(iii) Given: ⁿC₁₆ = ⁿC₁₄

To find: ⁿC₂₇=?

We know that:

ⁿC_r = ⁿC_n-r

⇒ ⁿC₁₄= ⁿC_n-14

⇒ ⁿC_n-14= ⁿC₁₆

⇒n-14=16

⇒n=16+14=30

⇒n=30

So,

ⁿC₂₇= ³⁰C₂₇

⇒³⁰C₂₇=

⇒³⁰C₂₇=

⇒³⁰C₂₇=

⇒³⁰C₂₇=

⇒³⁰C₂₇=

⇒³⁰C₂₇=4060

Ans: ³⁰C₂₇=4060

Given: ²⁰C_r = ²⁰C_r+6

To find: r=?

We know that:

ⁿC_r = ⁿC_n-r

⇒ ²⁰C_r+6= ²⁰C_20-(r+6)

⇒ ²⁰C_r+6= ²⁰C_20-r-6=²⁰C_14-r

⇒ ²⁰C_14-r= ²⁰C_r

⇒14-r=r

⇒2r=14

⇒

Ans:r=7

ii) Given: ¹⁸C_r = ¹⁸C_r+2

To find: ^rC₅=?

We know that:

ⁿC_r = ⁿC_n-r

⇒ ¹⁸C_r+2= ¹⁸C_18-(r+2)

⇒ ¹⁸C_r+2= ¹⁸C_18-r-2=¹⁸C_16-r

⇒ ¹⁸C_16-r= ¹⁸C_r

⇒16-r=r

⇒2r=16

⇒

So,

^rC₅= ⁸C₅

⇒⁸C₅=

⇒⁸C₅=

⇒⁸C₅=

⇒⁸C₅=

⇒⁸C₅=

⇒⁸C₅=56

Ans: ⁸C₅=56

Given: ⁿC_r–1 = ⁿC_3r

To find: r=?

We know that:

ⁿC_r = ⁿC_n-r

⇒ ⁿC_r–1 = ⁿC_n-(r-1)

⇒ ⁿC_r–1 = ⁿC_n-r+1

⇒ ⁿC_n-r+1= ⁿC_3r

⇒n-r+1=3r

⇒4r=n+1

⇒

Ans:

Given: ²ⁿC₃: ⁿC₃ = 12 : 1

To find: n=?

²ⁿC₃: ⁿC₃ = 12 : 1

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒2n-1=3(n-2)

⇒2n-1=3n-6

⇒n=6-1=5

Ans:n=5

Given: ¹⁵C_r : ¹⁵C_r–1 = 11 : 5

To find: r=?

¹⁵C_r : ¹⁵C_r–1 = 11 : 5

⇒

⇒

⇒

⇒

⇒

⇒5×(16-r)=11r

⇒80-5r=11r

⇒16r=80

⇒

⇒r=5

Ans:r=5

Given: ⁿP_r = 840 and ⁿC_r = 35

To find: r=?

We know that:

ⁿC_r =

and

ⁿP_r =

⇒ ⁿP_r= ⁿC_r×r!

⇒840=35×r!

⇒

⇒r!=4!

⇒r=4

Ans:r=4

Given, ⁿC_r–1 = 36, ⁿC_r = 84 and ⁿC_r+1 = 126

To find:r=?

⇒

⇒

⇒

⇒3(n-r+1)=7r

⇒3n-10r=-3….(1)

⇒

⇒

⇒

⇒2(n-r)=3(r+1)

⇒2n-5r=3….(2)

From equations 1 & 2 we get

n=9 & r=3

Ans:r=3

Given: ⁿ⁺¹C_r+1 : ⁿC_r = 11 : 6 and ⁿC_r : ^n–1C_r–1 = 6 : 3

To Find:n & r

We use this property in this question:

ⁿ⁺¹C_r+1 : ⁿC_r = 11 : 6

⇒

⇒

⇒

⇒6(n+1)=11(r+1)

⇒6n+6=11r+11

⇒6n-11r=5 …(1)

ⁿC_r : ^n–1C_r–1 = 6 : 3

⇒

⇒

⇒

⇒n=2r…(2)

Using equations 1 & 2 we get

⇒6(2r)-11r=5

⇒12r-11r=5

⇒r=5

⇒n=2×5

⇒n=10

Ans:n=10 & r=5

Condition: Each student has an equal chance of getting selected.

Imagine selecting the teammates one at a time. There are 15 ways of selecting the first teammate, 14 ways of selecting the second, 13 ways of selecting the third teammate, and so on down to 5 ways of selecting the eleventh teammate.

This is a problem of combination

⇒n=15 & r=11

⇒ ⁿC_r= ¹⁵C₁₁

⇒¹⁵C₁₁=

⇒¹⁵C₁₁=

⇒¹⁵C₁₁=

⇒¹⁵C₁₁=

⇒¹⁵C₁₁=1365

Ans: There can be 1365 different ways of choosing 11 players from a squad of 15.

This means there can be 1365 eleven-member teams formed with 15 players.

With 12 people , we need to choose a subset of two different people where order does not matter. Also, we need to choose all such subsets because each person is shaking hands with everyone else exactly once. The number of ways is: ⁿC_r

Where:n=12 & r=2

ⁿC_r= ¹²C₂

⇒¹²C₂=

⇒¹⁵C₁₁=

⇒¹⁵C₁₁=

⇒¹⁵C₁₁=

⇒¹⁵C₁₁=66

Ans:In total 66 handshakes are possible,if there are 12 persons in a party and if each two of them shake hands with each other

Number of points=21

⇒n=21

A chord connects circle at two points.

⇒r=2

⇒Number of chords from 21 points= ⁿC_r

⇒ ⁿC_r= ²¹C₂

⇒²¹C₂=

⇒²¹C₂=

⇒²¹C₂=

⇒²¹C₂=

⇒²¹C₂=210 chords.

Ans: 210 chords can be drawn through 21 points on a circle.

This is a case of combination:

Here,

n=25

r=4

⇒ ⁿC_r= ²⁵C₄

⇒²⁵C₄=

⇒²⁵C₄=

⇒²⁵C₄=

⇒²⁵C₄=

⇒²⁵C₄=12650 possible ways.

Ans:In 12650 ways ,from a class of 25 students, 4 can be chosen for a competition.

As there are 10 sportsmen out of which 5 are to be selected.

5 sportsmen can be selected out of 10 in ¹⁰C₅ ways.

Applying ⁿC_r =

We get,

⇒ ¹⁰C₅ =

252 ways

Hence, there are 252 ways of selecting 5 sportsmen from 10 sportsmen.

There are 5 black and 6 red balls. So,

The number of ways of selecting 2 black balls from 5 black balls is ⁵C_2, and number of ways of selecting 3 red balls from 6 red balls is ⁶C₃ .

Thus using the multiplication principle , the total number of ways will be

⁵C₂ × ⁶C₃ ways.

Applying ⁿC_r =

⇒200 ways

Thus, the total number of ways in which 2 black and 3 red balls can be selected is 200.

Total number of red balls = 6

Total number of white balls = 5

Total number of blue balls = 4

No. of ways of selecting 3 balls which is red = ⁶C₃

No. of ways of selecting 3 balls which is white = ⁵C₃

No. of ways of selecting 3 balls which is blue = ⁴C₃

Thus,by Multiplication principle, the total number of ways would be,

⁶C₃⁵C₃⁴C₃

Applying formula, ⁿC_r = , we get

⇒800 ways

Thus, the number of ways of selecting 9 balls from 6 red balls, 5 while balls and 4 blue balls if each selection consists of 3 balls of each colour would be 800.

Number of ways of choosing 5 boys out of 20 boys = ²⁰C₅

= 19 × 17 × 16 × 3 = 15,504

Number of ways of choosing 3 girls out of 10 girls = ¹⁰C₃

= 15 × 8 = 120

Total number of ways = 120 × 15,504 = 1,860,480

OR

Total number of ways = ²⁰C₅ × ¹⁰C₃

Since every student needs to choose 5 courses out of which 2 are compulsory. So, the student needs to choose 3 subjects out of 7 .

No. of ways for choosing 3 subjects out of 7 is ⁷C₃

Applying formula, ⁿC_r = , we get

= 35 ways.

There are 20 students in each classes and there is need of at least 5 students in each class to form a team of team of 11.

Now,

There are two ways in which the selection can be possible

1. Selecting 5 from XI and 6 from XII

2. Selecting 6 from XI and 5 from XII

Now, considering first case ,

No. of ways in selection of 5 students from 20 in class XI = ²⁰C₅

No. of ways in selection of 6 students from 20 in class XII = ²⁰C₆

By multiplication principle total no. of ways in first case is

= ²⁰C₅²⁰C₆

Now, considering second case,

No. of ways in selection of 6 students from 20 in class XI = ²⁰C₆

No. of ways in selection of 5 students from 20 in class XII = ²⁰C₅

By multiplication principle total no. of ways in second case is

= ²⁰C₆²⁰C₅

Now the total no. of ways will be the addition of both the cases

= = ²⁰C₅²⁰C₆ + ²⁰C₆²⁰C₅

= 2²⁰C₆²⁰C₅

Thus these are the ways by which A sports team of 11 students is to be constituted

The team of 6 has to be chosen from 4 officers and 8 clerks. There are some restrictions which are

1. To include exactly one officer

In this case ,

One officer will be chosen from 4 in ⁴C₁ ways

Therefore, 5 will be chosen form 8 clerks in ⁸C₅ ways.

Thus by multiplication principle , we get

Total no. of ways in 1 case is ⁴C₁⁸C_5.

2. To include at least one officer

In this case, there will be subcases for selection which is as follows.

(i) One officer and 5 clerks

(ii) Two officers and 4 clerks

(iii) Three officers and 3 clerks

(iv) Four officers and 2 clerks

Or

The required case of at least on officer would be

= Total cases – cases having only clerks

Now,

The total case would be choosing 6 out of 12 in ¹²C₆ ways.

And cases that would have only clerks would be i.e. selecting 6 from 8 clerks in ⁸C₆ ways.

¹²C₆ - ⁸C₆ ways.

Applying ⁿC_r =

⇒924 - 28 ways

= 896 ways

There is a cricket team of 11 players is to be selected from 16 players, which must include 3 bowlers and a wicketkeeper.

there will be a team of 7 batsmen, 1 wicketkeeper and 3 bowlers.

there are 5 bowlers from which 3 is to be selected in ⁵C₃ ways

⇒there are two wicketkeepers out of which 1 is to be selected in²C₁

hence, from 9 players left 7 is to be selected from that in ¹¹C₇ ways.

by Multiplication principle , we get

= ⁵C₃²C₁⁹C₇

Applying ⁿC_r =

= 720 ways

A team of 11 players is to be made from 25 players.

selecting 5 batsmen from 10 in ¹⁰C₅ ways.

selecting 3 all-rounders from 5 in ⁵C₃ ways.

selecting 2 bowlers from 8 in ⁸C₂ ways.

selecting 1 wicketkeeper from 2 in ²C₁ ways.

Thus , by the multiplication principle, we get

= ¹⁰C₅⁸C₂⁵C₃²C₁ ways

Applying ⁿC_r =

= 141120 ways

The question paper has two sets each containing 10 questions . so the student has to choose 8 from part A and 5 from part B.

⇒choosing 8 questions from 10 of part A in¹⁰C₈

⇒ choosing 5 questions from 10 of part B in¹⁰C₅

⇒by Multiplication principle, we get

=total no. of ways in which he can attempt the paper is¹⁰C₈¹⁰C₅

Applying ⁿC_r =

A student has to answer 4 questions out of 5 in which he is compelled to do the 1 and 2 questions compulsory. So he has to attempt 2 questions from 3 of his choice.

Choosing 2 questions from 3 will be in ³C₂ ways.

Applying ⁿC_r =

= 3 ways.

There are total 13 questions out of which 10 is to be answered .The student can answer in the following ways:

6 questions from part A and 4 from part B

5 questions from part A and 5 from part B

4 questions from part A and 6 from part B

⇒total ways in the 1st case are ⁶C₆⁷C₄

⇒ total ways in the 2nd case are ⁶C₅⁷C₅

⇒ total ways in the 3rd case are ⁶C₄⁷C₆

= ⁶C₆⁷C₄+⁶C₅⁷C₅⁶C₄⁷C₆

Applying ⁿC_r =

= 266 ways.

There are total 13 questions out of which 10 is to be answered .The student can answer in the following ways:

3 questions from part A and 4 from part B

4 questions from part A and 3 from part B

5 questions from part A and 2 from part B

2 questions from part A and 5 from part B

⇒total ways in the 1st case are ⁶C₃⁶C₄

⇒ total ways in the 2nd case are ⁶C₄⁶C₃

⇒ total ways in the 3rd case are ⁶C₅⁶C₂

⇒ total ways in the 4th case are ⁶C₂⁶C₅

=⁶C₄⁶C₃+⁶C₃⁶C₄⁶C₅⁶C₂+ ⁶C₂⁶C₅

Applying ⁿC_r =

= 780 ways.

Since the committee of 11 is to be formed from 6 teachers and 8 students.

(i) Forming a committee with exactly 4 teachers

Choosing 4 teachers out of 6 in ⁶C₄ ways.

Remaining 7 from 8 students in ⁸C₇ ways.

Thus, total ways in (i) are ⁶C₄⁸C₇ ways.

(ii) The number of ways in this case is

1. 4 teachers and 7 students

2. 5 teachers and 6 students

3. 6 teachers and 5 students

= ⁶C₄⁸C₇+⁶C₅⁸C₆⁶C₆⁸C₅

Applying ⁿC_r =

= 344 ways

A committee of 7 has to be formed from 9 boys and 4 girls.

I. Exactly 3 girls: If there are exactly 3 girls in the committee, then there must be 4 boys, and the ways in which they can be chosen is

= ⁴C₃⁹C₄

= 504 ways

II. At least 3 girls: Here the possibilities are

(i) 3 girls and 4 boys and

(ii) 4 girls and 3 boys.

the number of ways they can be selected

= ⁴C₃⁹C₄ + ⁴C₄⁹C₃

₌ 588

III. At most 3 girls:

(i) 7 boys but no girls

(ii) 6 boys and 1 girl

(iii) 5 boys and 2 girls &

(iv) 4 boys and 3 girls.

And the number of their selection is

= ⁴C₃⁹C₄ + ⁴C₂⁹C₅ + ⁴C₁⁹C₆ + ⁴C₀⁹C₇

= 1584 ways.

Total number of persons = 2 + 3 = 5

Now, committee consist of 3 persons.

Therefore, total number of ways = ⁵C₃ = 5 × 2 = 10

Now,

When 1 man is selected, total ways = ²C₁

When 2 women are selected, total ways = ³C₂

Total number of ways when 1 man and 2 women are selected = ²C₁ × ³C₂ = 2 × 3 = 6

Since the committee of 5 is to be formed from 6 gents and 4 ladies.

(i) Forming a committee with at least 2 ladies

Here the possibilities are

(i) 2 ladies and 3 gents

(ii) 3 ladies and 2 gents

(iii) 4 ladies and 1 gent

the number of ways they can be selected

= ⁴C₂⁶C₃ + ⁴C₃⁶C₂ + ⁴C₄⁶C₁

Applying ⁿC_r =

= 186 ways

(ii) The number of ways in this case is

1. 0 ladies and 5 gents

2. 1 lady and 4 gents

3. 2 ladies and 3 gents.

The total ways are

= ⁴C₀⁶C₅ + ⁴C₁⁶C₄ + ⁴C₂⁶C₃

Applying ⁿC_r =

= 186 ways.

2 girls who won the prize last year are surely to be taken. So, we have to make a selection of 8 students out of 14 boys and 8 girls, choosing at least 4 boys and at least 2 girls.

Thus, we may choose:

(4 boys, 4 girls) or (5 boys, 3 girls) or (6 boys, 2 girls)

Therefore, the required number of ways = (¹⁴C₄ × ⁸C₄) + (¹⁴C₅ × ⁸C₃) + (¹⁴C₆ × ⁸C₂)

Since there are 52 cards in a deck out of which 4 are king and others are non-kings.

So, the no. of ways are as follows:

1. 1 king and 4 non-king

2. 2 king and 3 non-king

3. 3 king and 2 non-king

4. 4 king and 1 non-king

So , total no. of ways are

= ⁴C₁⁴⁸C₄ + ⁴C₂⁴⁸C₃ + ⁴C₃⁴⁸C₂ + ⁴C₄⁴⁸C₁

Applying ⁿC_r =

= 778320+103776+4512+48

= 886656 ways.

For a diagonal to be formed, 2 vertices are required. Thus in a polygon, there are 10 sides. And no. of lines can be formed are ⁿC_2, but in ⁿC₂ the sides are also included. N of them is sides.

Thus the no. of diagonals are ⁿC₂ - n

(i) hexagon

n=6

so no of diagonal is ⁶C₂-6
= 9

(ii) decagon

n= 10

so no of diagonal is ¹⁰C₂-10

= 35

(iii) N= 18

so no of diagonal is ¹⁸C₂-18

= 135

Total number of points on plane = 12

Triangles can be formed from these points = ¹²C₃

= 220

But 4 points are colinear, the number of triangles can be formed from these points

= ⁴C₃

= 4

We need to subtract 4 from 220 because in the formation of triangles from 4 colinear points are added there.

So no of triangle formed is = 220-4

= 216

Total number of sides in a decagon = 10

We know that number of vertices in triangle = 3

So, out of 10 vertices we have to choose 3 vertices.

Therefore,

Total number of triangles in a decagon = ¹⁰C₃

Total number of triangles = 120

Since there are 10 different books out of which 4 is to be selected .

(i) When there is no restriction

No. of ways in which 4 books be selected = ¹⁰C₄

= 210 ways

(ii) two particular books are always selected

since two particular books are always selected, so ways of selecting 2 books from 8 are = ⁸C₂ ways

= 28 ways

(iii) two particular books are never selected

since two particular books are never selected so , ways of selecting 4 books from 8 are = ⁸C₄ ways

= 70 ways

The given no. is 3,5,7,11.

The no. of different products when two or more is taking= the no. of ways of taking the product of two no.+ the no. of ways of taking the product of three no. + the no. of ways of taking the product of four no.

= ⁴C₂ + ⁴C₃ + ⁴C₄

Applying ⁿC_r =

= 6+4+1

=11

Since a committee is to be formed of 2 teachers and 3 students

(i) When a particular teacher is included

No. of ways in which committee can be formed = ⁹C₁²⁰C₃

= 9720 ways

(ii) a particular student is included

since a particular student is always selected so ways of selecting 2 teachers and 2 students from 10 and 19 respect. is = ¹⁰C₂¹⁹C₂ ways

= 7695 ways

(iii) a particular student is excluded

since 1 particular student is excluded so , ways of selecting 2 teachers and 3 students from 10 and 19 respt. is = ¹⁰C₂¹⁹C₃ ways

= 43605 ways

A line is formed by joining two points.

If the total number of points is 18, the total number of lines would be = ¹⁸C₂

But 5 points are collinear, so the lines made by these points are the same and would be only 1.

Hence there is 1 common line joining the 5 collinear points.

As these 5 points are also included in 18 points so these must be subtracted from the total case , i.e. ⁵C₂ must be subtracted from ¹⁸C_{2 .}

Finally, the number of straight line = ¹⁸C₂ - ⁵C₂ + 1

= 144 lines

3 consonants out of 12 consonants can be chosen in ¹²C₃ ways. 2 vowels out of 5 vowels can be chosen in ⁵C₂ ways. And also 5 letters can be written in 5! Ways. Therefore, the number of words can be formed is (¹²C₃ X⁵C₂ X 5!) = 264000.

In the word ‘INVOLUTE’ there are 4 vowels, ‘I’,’O’,’U’ and ‘E’ and there are 4 consonants, ‘N’,’V’,’L’ and ‘T’. 3 vowels out of 4 vowels can be chosen in ⁴C₃ ways. 2 consonants out of 4 consonants can be chosen in ⁴C₂ ways. Length of the formed words will be (3 + 2) = 5. So, the 5 letters can be written in 5! Ways. Therefore, the total number of words can be formed is = (⁴C₃ X ⁴C₂ X 5!) = 2880.

2 consonants out of 21 consonants can be chosen in ²¹C₂ ways. 3 vowels out of 5 vowels can be chosen in ⁵C₃ ways. Length of the word is = (2 + 3) = 5 And also 5 letters can be written in 5! Ways. Therefore, the number of words can be formed is = (²¹C₂ X ⁵C₃ X 5!) = 252000.

The seating arrangement would be like this: B G B G B G B G B So, 4 girls can seat among the four places. Number of ways they can seat is = ⁴P₄ = 24 Boys have to seat among the ‘B’ areas. So, there are 5 seats available for 3 boys. The number of ways the 3 boys can seat among the 5 places is = ⁵P₃ = 60 Therefore, the total number of ways they can seat in this manner is = (24 X 60) = 1440.

There are 6 letters in the word ‘MONDAY’, and there is no letter repeating. (i) 4 letters are used at first. 4 letters can sit in different ways. So, here permutation is to be used. So, the number of words that can be formed = ⁶P₄ = 360. [Answer(i)] (ii) Now all the letters are used. Therefore, the number of words can be formed is = 6! = 720 [Answer(ii)] (iii) Now the first letter is a vowel. There are 2 vowels in the word ‘MONDAY’, ‘O’ and ‘A’. Let’s take ‘O’ as the first letter. Then we can place the 5 letters among the 5 places. So, taking ‘O’ as the first letter, a number of words can be formed is = 5! = 120. Similarly, taking ‘A’ as the first letter, a number of words can be formed = 5! = 120. So, the total number of words can be formed taking first letter a vowel is = (120 + 120) = 240. [Answer(iii)]

Given: ²⁰C_r = ²⁰C_r–10 Need to find: Value of ¹⁷C_r We know, one of the property of combination is: If ⁿC_r = ⁿC_t, then, (i) r = t OR (ii) r + t = n We can’t apply the property (i) here. So we are going to use property (ii) ²⁰C_r = ²⁰C_r–10 By the property (ii), ⇒ r + r – 10 = 20 ⇒ 2r = 30 ⇒ r = 15 Therefore, ¹⁷C₁₅ = 136.

Given: ²⁰C_r+1 = ²⁰C_r–10 Need to find: Value of ¹⁰C_r We know, one of the property of combination is: If ⁿC_r = ⁿC_t, then, (i) r = t OR (ii) r + t = n We can’t apply the property (i) here. So we are going to use property (ii) ²⁰C_r+1 = ²⁰C_r–10 By the property (ii), ⇒ r + 1 + r – 10 = 20 ⇒ 2r = 29 ⇒ r = 14.5 We need to find out the value of ¹⁰C_r. But here r can’t be a rational number. Therefore the value of ¹⁰C_r can’t be find out.

Given: ⁿC_r+1 = ⁿC₈ Need to find: Value of ²²C_n We know, one of the property of combination is: If ⁿC_r = ⁿC_t, then, (i) r = t OR (ii) r + t = n We are going to use property (i) ⁿC_r+1 = ⁿC₈ By the property (i), ⇒ r + 1 = 8 ⇒ r = 7 Now we are going to use property (ii) ⇒ n = 8 + 7 + 1 = 16 Therefore, ²²C_n = ²²C₁₆ = 74613.

Given: ³⁵C_n+7 = ³⁵C_4n–2 Need to find: Value of n We know, one of the property of combination is: If ⁿC_r = ⁿC_t, then, (i) r = t OR (ii) r + t = n Applying property (i) we get, ⇒ n + 7 = 4n – 2 ⇒ 3n = 9 ⇒ n = 3 Applying property (ii) we get, ⇒ n + 7 + 4n – 2 = 35 ⇒ 5n = 30 ⇒ n = 6 Therefore, the value of n is either 3 or 6.

(i) ²⁰⁰C₁₉₈(ii) ⁷⁶C₀ [As 0! = 1] (iii) ¹⁵C₁₅

Given: ^mC₁ = ⁿC₂ Need to prove: m = n(n – 1) ^mC₁ = ⁿC₂ ⇒ ⇒ ⇒ [Proved]

⇒ ⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅ ⇒ ⁶C₂ + ⁶C₄ + 1 [As ⁵C₅ = 1] ⇒ 15 + 15 + 1 ⇒ 31

Given: ⁿ⁺¹C₃ = 2(ⁿC₂) Need to find: Value of n ⇒ ⁿ⁺¹C₃ = 2(ⁿC₂) ⇒ ⇒ ⇒ [As ]

Given: ⁿP_r = 720 & ⁿC_r = 120 Need to find: Value of r We know that, ⁿP_r = r! X ⁿC_r Putting the values, ⇒ 720 = r! X 120 ⇒ r! = 6 = 3! ⇒ r = 3.

Given: ^(n2–n)C₂ = ^(n2–n)C₄ = 120 Need to find: Value of n ^(n2–n)C₂ = ^(n2–n)C₄ = 120 We know, one of the property of combination is: If ⁿC_r = ⁿC_t, then, (i) r = t OR (ii) r + t = n Applying property (ii) we get, n² – n = 2 + 4 = 6 n² – n – 6 = 0 n² – 3n + 2n – 6 = 0 n(n – 3) + 2(n – 3) = 0 (n – 3) (n + 2) = 0 So, the value of n is either 3 or -2.

3 consonants out of 5 consonants can be chosen in ⁵C₃ ways. 2 vowels out of 4 vowels can be chosen in ⁴C₂ ways. And also 5 letters can be written in 5! Ways. Therefore, the number of words can be formed is (⁵C₃ X⁴C₂ X 5!) = 7200.

n-sided polygon has n numbers of vertices. Diagonals are formed by joining the opposite vertices from one vertex, except the two adjacent vertices. So, from one vertex (n-3) diagonals can be drawn. Similarly, for n numbers of vertices, n(n-3) diagonals can be drawn. But, the diagonal joins 2 points at a time, here two vertices. Therefore, the actual number of diagonals is = .

Three persons enter a compartment where 5 seats are vacant. The number of ways they can be seated is = ⁵P₃ = 60.

To get a straight line we just need to join two points. There are 12 numbers of points. Therefore, there is ¹²C₂ = 66 number of straight lines. Among the 12 points, there are 3 points which are collinear. That means joining those 3 lines give a single straight line. That means the real number of straight lines present in the table is = (66 – ³C₂ + 1) = (66 – 3 + 1) = 64.

We need to include at least 2 women. If we include 2 women in the committee, then a number of men is 3. The number of ways, 2 women can be selected out of 4 is = ⁴C₂ = 6 The number of ways, 3 men can be selected out of 6 is = ⁶C₃ = 20 So, the committee can be formed including 2 women in (20 X 6) = 120 ways. If we include 3 women in the committee, then a number of men is 2. The number of ways, 3 women can be selected out of 4 is = ⁴C₃ = 4 The number of ways, 2 men can be selected out of 6 is = ⁶C₂ = 15 So, the committee can be formed including 3 women in (15 X 4) = 60 ways. Therefore, the total number of ways the committee can be formed is = (120 + 60) = 180 ways.

There are 4 bowlers in 13 player team. So, maximum we can add 4 bowlers. And we need to include at least 3 bowlers. If we include 3 bowlers then from the remaining 9 [13 – 4 bowlers] players, we need to include 8. The number of ways, 8 players can be selected among 9 is = ⁹C₈ = 9 The number of ways, 3 players can be selected among 4 is = ⁴C₃ = 4 So, taking 3 bowlers the team can be represented in (9 X 4) = 36 ways. If we include 4 bowlers then from the remaining 9 [13 – 4 bowlers] players, we need to include 7. The number of ways, 7 players can be selected among 9 is = ⁹C₇ = 36 The number of ways, 4 players can be selected among 4 is = ⁴C₄ = 1 So, taking 4 bowlers the team can be represented in (36 X 1) = 36 ways. Therefore, the total possible ways are = (36 + 36) = 72.

Each committee consists of 3 men and 2 women. So, we need to select 3 men out of 6 and 2 women out of 4. The number of ways, 3 men can be selected out of 6, is = ⁶C₃ = 20 The number of ways, 2 women can be selected out of 4, is = ⁴C₂ = 6 So, the totally (20 + 6) = 26 numbers of different committees can be formed.

To form a parallelogram we need 2 sets of 2 parallel lines intersecting the other 2 lines from the other set. So, first of all, we need to get 2 lines from the sets. From the first parallel set, 2 out of 4 lines can be selected in ⁴C₂ = 6 ways. From the second parallel set, 2 out of 3 lines can be selected in ³C₂ = 3 ways. So, the total number of parallelograms can be formed is = (6 X 3) = 18.